dx-x-16-4x-2-

Question Number 141943 by cesarL last updated on 25/May/21

\int \frac{d x}{x \sqrt{16 - 4 x^{2}}}

Answered by MJS_new last updated on 25/May/21

\int \frac{d x}{x \sqrt{16 - 4 x^{2}}} = \frac{1}{2} \int \frac{d x}{x \sqrt{4 - x^{2}}} =

[t = \frac{2 + \sqrt{4 - x^{2}}}{x} \to d x = - \frac{x^{2} \sqrt{4 - x^{2}}}{2 (2 + \sqrt{4 - x^{2}})} d t]

= - \frac{1}{4} \int \frac{d t}{t} = - \frac{1}{4} \ln t =

= \frac{1}{4} \ln ∣ x ∣ - \frac{1}{4} \ln (2 + \sqrt{4 - x^{2}}) + C

Commented by cesarL last updated on 25/May/21

I n e e d w i t h t r i g o n o m e t r i c s u s t i t u t i o n

Commented by Ar Brandon last updated on 25/May/21

Then you may let x = 2 \sin θ

Answered by mathmax by abdo last updated on 25/May/21

Ψ = \int \frac{dx}{x \sqrt{16 - {4 x}^{2}}} \Rightarrow Ψ = \int \frac{dx}{2 x \sqrt{4 - x^{2}}} [changement x = 2 sint give

Ψ = \int \frac{2 cost}{4 sint . 2 cost} dt = \frac{1}{4} \int \frac{dt}{sint} =_{\tan (\frac{t}{2}) = y} \frac{1}{4} \int \frac{2 dy}{(1 + y^{2}) \frac{2 y}{1 + y^{2}}}

= \frac{1}{4} \int \frac{dy}{y} = \frac{1}{4} \log ∣ y ∣ + C = \frac{1}{4} \log ∣ \tan \frac{t}{2} ∣ + C

t = \arcsin (\frac{x}{2}) \Rightarrow Ψ = \frac{1}{4} \log ∣ \tan (\frac{1}{2} \arcsin (\frac{x}{2}) ∣ + C